Henri Mouhot
(1826-1861)

French explorer – Henri Mouhot was born in Montbéliard (Commune in Doubs, France) on May 15th, 1826 and died in Luang Prabang (City located in north central Laos) on November 10th, 1861 at the age of 35. Today Henri Mouhot would be 197 years old.

Age

how old was Henri Mouhot when he died?

35

Henri Mouhot died in 1861 at the age of 35.


Biographical data

Birthday May 15th, 1826 (Monday)
Place of Birth Montbéliard
Commune in Doubs, France
Death Date November 10th, 1861 (Sunday)
Death place Luang Prabang
City located in north central Laos
Birth sign (Zodiac) Taurus (The Bull) ♉
Chinese Zodiac Dog 狗

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Henri Mouhot

Birthday

When was Henri Mouhot born?

Henri Mouhot was born on May 15th, 1826.

Death

When did Henri Mouhot die?

Henri Mouhot died on November 10th, 1861 at the age of 35 in Luang Prabang (City located in north central Laos). Today Henri Mouhot would be 197 years old.

Place of Birth

Where was Henri Mouhot born?

Montbéliard (Commune in Doubs, France).

Death place

Where did Henri Mouhot die?

Luang Prabang (City located in north central Laos).

Zodiac

What is the zodiac sign of Henri Mouhot?

Henri Mouhot was born in the zodiac sign Taurus (The Bull).

Chinese Zodiac

Which Chinese zodiac sign does Henri Mouhot have?

Henri Mouhot was born in 1826 in the year of the Chinese zodiac Dog.

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Died on November 10th, 1861:
Anniversaries of death

French explorer

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